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\title{{\Huge Homework6\\ 数值分析}}
\author{{\LARGE 3190105815 信息与计算科学\ 行一凡}}
\date{{\LARGE \today}}

\begin{document}
	\maketitle
	\section*{Exercise I}
	\begin{enumerate}[(a)]
		\item 计算差分表得到
		\begin{center}
			\begin{tabular}{c|cccc}
				$ -1 $ & $ y(-1) $ & & &\\ 
				$ 0 $ & $ y(0) $ & $ y(0)-y(-1) $ & &\\ 
				$ 0 $ & $ y(0) $ & $ y^{\prime}(0) $ & $ y^{\prime}(0)-y(0)-y(-1) $ &\\ 
				$ 1 $ & $ y(1) $ & $ y(1)-y(0) $ & $ y(1)-y(0)-y^{\prime}(0) $ & $ \frac{1}{2}\left[y(1)-2y^{\prime}(0)-y(-1)\right] $ 
			\end{tabular}
		\end{center}
		从而有插值多项式
		\begin{equation}\label{key}
			\begin{aligned}
				p_3(t) &= y(-1)+[y(0)-y(-1)](t+1)+[y^{\prime}(0)-y(0)-y(-1)]t(t+1)\\ &+\frac{1}{2}\left[y(1)-2y^{\prime}(0)-y(-1)\right]t^{2}(t+1)
			\end{aligned}
		\end{equation}
		\begin{equation}\label{key}
			\begin{aligned}
				\int_{-1}^{1}p_3(t)dt &= 2y(-1)+2[y(0)-y(-1)]+\frac{2}{3}[y^{\prime}(0)-y(0)-y(-1)]\\ 
				&+\frac{1}{3}\left[y(1)-2y^{\prime}(0)-y(-1)\right]\\ 
				&= \frac{1}{3}[y(-1)+4y(0)+y(1)]\\ 
				&= I_S(y)
			\end{aligned}
		\end{equation}
	
		\item 由 Hermite 插值问题的余项，运用积分中值定理
		\begin{equation}\label{key}
			\begin{aligned}
				E_S(y) &= -\int_{-1}^{1}\dfrac{f^{4}(\xi(x))}{4!}t^{2}(t+1)(1-t)dt\\ 
				&= -\dfrac{f^{(4)}(\zeta)}{4!}\int_{-1}^{1}t^{2}(t+1)(1-t)dt\\ 
				&= -\dfrac{f^{(4)}(\zeta)}{90}
			\end{aligned}
		\end{equation}
	
		\item 利用变换$ g:[-1,1]\to[a,b] $，定义
		\begin{equation}\label{key}
			g(t) = \dfrac{b-a}{2}t + \dfrac{b+a}{2}
		\end{equation}
		则有
		\begin{equation}\label{key}
			\int_{a}^{b}p_3(t)dt = \int_{-1}^{1}p_3(g(t))g^{\prime}(t)dt = \int_{-1}^{1}\dfrac{b-a}{2}p_3\left(\dfrac{b-a}{2}t + \dfrac{b+a}{2}\right)dt
		\end{equation}
		定义
		\begin{equation}\label{key}
			h(t) = \dfrac{b-a}{2}p_3\left(\dfrac{b-a}{2}t + \dfrac{b+a}{2}\right)
		\end{equation}
		\begin{equation}\label{key}
			z(t) = \dfrac{b-a}{2}y\left(\dfrac{b-a}{2}t + \dfrac{b+a}{2}\right)
		\end{equation}
		则可得
		\begin{equation}\label{key}
			\begin{aligned}
				&h(-1) = \dfrac{b-a}{2}y(a) = z(-1),\quad h(0) = \dfrac{b-a}{2}y\left(\dfrac{b+a}{2}\right) = z(0),\\ 
				&h^{\prime}(0) = \left(\dfrac{b-a}{2}\right)^{2}y^{\prime}\left(\dfrac{b+a}{2}\right) = z^{\prime}(0),\quad h(1) = \dfrac{b-a}{2}y(b) = z(1)
			\end{aligned}
		\end{equation}
		由(a)得
		\begin{equation}\label{key}
			\begin{aligned}
				\int_{a}^{b}p_3(t)dt &= \int_{-1}^{1}h(t)dt = \dfrac{1}{3}[z(-1)+4z(0)+z(1)]\\ 
				&= \dfrac{b-a}{6}\left[y(a)+4y\left(\dfrac{b+a}{2}\right)+y(b)\right]
			\end{aligned}
		\end{equation}
		应用到对$ [a,b] $等分的区间上，$ h = \frac{b-a}{n} $，则上面的$ b-a $替换为$ 2h $，有
		\begin{equation}\label{key}
			I_n^{S}(f) = \dfrac{h}{3}\sum_{i=1}^{\frac{n}{2}}[f(x_{2i-2})+4f(x_{2i-1})+f(x_{2i})]
		\end{equation}
		考虑 Hermite 差插值问题的余项，设$ d = \frac{b-a}{2} $，运用积分中值定理
		\begin{equation}\label{key}
			\begin{aligned}
				E_S(f) &= -\int_{a}^{b}\dfrac{f^{4}(\xi(x))}{4!}\left(t-\dfrac{b+a}{2}\right)^{2}(t-a)(b-t)dt\\ 
				&= -\dfrac{f^{(4)}(\zeta)}{4!}\int_{a}^{b}\left(t-\dfrac{b+a}{2}\right)^{2}(t-a)(b-t)dt\\ 
				&= -\dfrac{f^{(4)}(\zeta)}{4!}\int_{-d}^{d}t^{2}(t+d)(d-t)dt\\ 
				&= -\dfrac{f^{(4)}(\zeta)}{90}d^{5} = -\dfrac{(b-a)^{5}}{2880}f^{(4)}(\zeta)
			\end{aligned}
		\end{equation}
		同样应用到对$ [a,b] $等分的区间上，有
		\begin{equation}\label{key}
			E_S(f) = -\dfrac{n}{2}\dfrac{(2h)^{5}}{2880}f^{(4)}(\zeta) = -\dfrac{b-a}{180}h^{4}f^{(4)}(\zeta)
		\end{equation}
	\end{enumerate}
	
	\section*{Exercise II}
	\begin{enumerate}[(a)]
		\item $ f(x) = e^{-x^{2}} $，由 composite trapezoidal rule 的余项公式有
		\begin{equation}\label{key}
			E_{n}^{T}(f) = -\dfrac{1}{12}h^2f^{\prime\prime}(\xi) = -\dfrac{1}{12}h^2(4\xi^{2}-2)e^{-\xi^{2}}
		\end{equation}
		\begin{equation}\label{key}
			f^{(3)}(\xi) = (12\xi-8\xi^3)e^{-\xi^{2}} \ge 0
		\end{equation}
		代入$ \xi = 0 $，从而有估计
		\begin{equation}\label{key}
			\left|E_{n}^{T}(f)\right| \le \dfrac{h^2}{6} = \dfrac{1}{6n^2}\le 0.5\times 10^{-6}
		\end{equation}
		得到$ n \approx 578 $.
	
		\item 由 composite Simpson's rule 的余项公式有
		\begin{equation}\label{key}
			E_{n}^{T}(f) = -\dfrac{1}{180}h^4f^{(4)}(\xi) = -\dfrac{1}{180}h^4(12-48\xi^{2}+16\xi^{4})e^{-\xi^{2}}
		\end{equation}
		\begin{equation}\label{key}
			f^{(5)}(\xi) = (-120\xi+160\xi^3-32\xi^{5})e^{-\xi^{2}}
		\end{equation}
		代入$ \xi = 0 $，从而有估计
		\begin{equation}\label{key}
			\left|E_{n}^{T}(f)\right| \le \dfrac{h^4}{15} = \dfrac{1}{15n^4}\le 0.5\times 10^{-6}
		\end{equation}
		得到$ n \approx 20 $.
	\end{enumerate}

	\section*{Exercise III}
	\begin{enumerate}[(a)]
		\item 需要$ \left<1,\pi_2\right> = \left<t,\pi_2\right> = 0, $，则有
		\begin{equation}\label{key}
			\begin{aligned}
				&\left<1,\pi_2\right> = \int_{0}^{+\infty}(t^2+at+b)e^{-t}dt = 2 + a + b\\ 
				&\left<t,\pi_2\right> = \int_{0}^{+\infty}t(t^2+at+b)e^{-t}dt = 6 + 2a + b
			\end{aligned}
		\end{equation}
		解得$ a = -4, b = 2 $，即有
		\begin{equation}\label{key}
			\pi_2(t) = t^2 - 4t + 2
		\end{equation}
	
		\item $ \pi_2(t) $有零点
		\begin{equation}\label{key}
			t_1 = 2 + \sqrt{2},\ t_2 = 2 - \sqrt{2}
		\end{equation}
		利用 Guass quadrature 对常数和线性多项式成立，
		\begin{equation}\label{key}
			\begin{aligned}
				&\omega_1 + \omega_2 = \int_{0}^{+\infty}e^{-t}dt = 1\\ 
				&t_1\omega_1 + t_2\omega_2 = \int_{0}^{+\infty}te^{-t}dt = 1
			\end{aligned}
		\end{equation}
		从而有
		\begin{equation}\label{key}
			\omega_1 = \dfrac{2-\sqrt{2}}{4},\ \omega_2 = \dfrac{2+\sqrt{2}}{4}
		\end{equation}
		\begin{equation}\label{key}
			I_2^{G}(f) = \left(\dfrac{2-\sqrt{2}}{4}\right)f\left(2 + \sqrt{2}\right) + \left(\dfrac{2+\sqrt{2}}{4}\right)f\left(2 - \sqrt{2}\right)
		\end{equation}
		由余项公式有
		\begin{equation}\label{key}
			E_2^{G}(f) = \dfrac{f^{(4)}(\tau)}{4!}\int_{0}^{+\infty}e^{-t}(t^2 - 4t + 2)^{2}dt = \dfrac{f^{(4)}(\tau)}{6}
		\end{equation}
	
		\item $ f(t) = \frac{1}{1+t} $代入估计公式有
		\begin{equation}\label{key}
			I_2^{G}(f) \approx 0.571428571
		\end{equation}
		由余项表达式仅能得到估计
		\begin{equation}\label{key}
			\left|E_2^{G}(f)\right| = \dfrac{6}{(1+\tau)^{5}} \le 6
		\end{equation}
		由精确值$ I = 0.596347361 $，真实误差为0.024918789，代入得
		\begin{equation}\label{key}
			\dfrac{6}{(1+\tau)^{5}} \approx 0.024918789
		\end{equation}
		则有$ \tau \approx 1.994503751 $.
	\end{enumerate}

	\section*{Exercise IV}
	\begin{enumerate}[(a)]
		\item 令$ h_m(t) $在$ x_1,\cdots, x_n $处满足
		\begin{equation}\label{key}
			h_m(x_i) = \left\{
			\begin{aligned}
				&1,\ i = m\\ 
				&0,\ i \neq m
			\end{aligned}
			\right.\quad
			h_m^{\prime}(x_i) = 0,\ i = 1,\cdots,n
		\end{equation}
		且有
		\begin{equation}\label{key}
			\begin{aligned}
				l_m(t) &= \prod_{i = 1\atop i \neq m}^{n}\dfrac{t-x_i}{x_m-x_i}\\ l_m^{\prime}(t) &= \sum_{i = 1\atop i \neq m}^{n}\left(\dfrac{1}{x_m-x_i}\prod_{j = 1\atop j \neq m,i}^{n}\dfrac{t-x_j}{x_m-x_j}\right)
			\end{aligned}
		\end{equation}
		显然有
		\begin{equation}\label{key}
			l_m(x_i) = \left\{
			\begin{aligned}
				&1,\ i = m\\ 
				&0,\ i \neq m
			\end{aligned}
			\right.\quad
			l_m^{\prime}(x_m) = \sum_{i = 1\atop i \neq m}^{n}\dfrac{1}{x_m-x_i}
		\end{equation}
		由$ h_m(t) = (a_m+b_mt)l_m^{2}(t) $得
		\begin{equation}\label{key}
			h_m^{\prime}(t) = b_ml_m^{2}(t) + 2(a_m+b_mt)l_m^{\prime}(t)l_m(t)
		\end{equation}
		从而有
		\begin{equation}\label{key}
			h_m(x_i) = \left\{
			\begin{aligned}
				&a_m+b_mx_m,\ &i = m\\ 
				&0,\ &i \neq m
			\end{aligned}
			\right.\quad
			h_m^{\prime}(x_i) = \left\{
			\begin{aligned}
				&b_m+2(a_m+b_mx_m)l_m^{\prime}(x_m),\ &i = m\\ 
				&0,\ &i \neq m
			\end{aligned}
			\right.
		\end{equation}
		只需求解线性方程组
		\begin{equation}\label{key}
			\left\{
			\begin{aligned}
				&a_m+b_mx_m = 1\\ 
				&b_m+2(a_m+b_mx_m)l_m^{\prime}(x_m) = 0
			\end{aligned}
			\right.
		\end{equation}
		解得
		\begin{equation}\label{key}
			a_m = 1 + \sum_{i = 1\atop i \neq m}^{n}\dfrac{2x_m}{x_m-x_i},\ b_m = -\sum_{i = 1\atop i \neq m}^{n}\dfrac{2}{x_m-x_i}
		\end{equation}
		类似地，要求
		\begin{equation}\label{key}
			q_m(x_i) = 0,\ i = 1,\cdots,n\quad
			q_m^{\prime}(x_i) = \left\{
			\begin{aligned}
				&1,\ i = m\\ 
				&0,\ i \neq m
			\end{aligned}
			\right.
		\end{equation}
		得到线性方程组
		\begin{equation}\label{key}
			\left\{
			\begin{aligned}
				&c_m+d_mx_m = 0\\ 
				&d_m+2(c_m+d_mx_m)l_m^{\prime}(x_m) = 1
			\end{aligned}
			\right.
		\end{equation}
		解得
		\begin{equation}\label{key}
			c_m = -x_m,\ d_m = 1
		\end{equation}
	
		\item 直接对(a)中所得$ p(x) $积分得到
		\begin{equation}\label{key}
			I_n(f) = \sum_{k=1}^{n}\left[\omega_kf(x_k)+\mu_kf^{\prime}(x_k)\right]
		\end{equation}
		其中
		\begin{equation}\label{key}
			\omega_k = \int_{a}^{b}\rho(t)h_m(t)dt,\ \mu_k = \int_{a}^{b}\rho(t)q_m(t)dt
		\end{equation}
		由于$ p(t) $实际上是$ f(x) $的 Hermite 插值多项式，根据 Hermite 插值的余项，对于$ f\in\mathbb{P}_{2n-1} $，有$ f(t) = p(t) $，从而$ E_n(f) = 0 $.
		
		\item 由(b)所得，$ \mu_k = 0, \forall k=1,\cdots,n $，有
		\begin{equation}\label{key}
			\int_{a}^{b}\rho(t)q_k(t)dt = 0,\ k=1,\cdots,n
		\end{equation}
		节点多项式为$ v_n(x) = \prod_{k=1}^{n}(x-x_k) $，则
		\begin{equation}\label{key}
			\begin{aligned}
				q_k(t) &= (t-x_k)\left(\prod_{i = 1\atop i \neq k}^{n}\dfrac{t-x_i}{x_k-x_i}\right)^{2}\\ 
				&= v_n(t)l_k(t)\prod_{i = 1\atop i \neq k}^{n}\dfrac{1}{x_k-x_i}
			\end{aligned}
		\end{equation}
		代入得
		\begin{equation}\label{key}
			\int_{a}^{b}\rho(t)v_n(t)l_k(t)dt = 0,\ k=1,\cdots,n
		\end{equation}
		则节点多项式满足$ \left<v_n, l_k\right> = 0,\ k=1,\cdots,n $，即节点多项式与任意初等 Lagrange 多项式正交.
	\end{enumerate}

	\section*{Exercise V}
	构造 FD 公式
	\begin{equation}\label{key}
		D^{2}u(\overline{x}) = au(\overline{x}-h) + bu(\overline{x}) + cu(\overline{x}+h)
	\end{equation}
	在$ x = \overline{x} $处 Taylor 展开得
	\begin{equation}\label{key}
		\begin{aligned}
			D^{2}u(\overline{x}) &= (a + b + c)u(\overline{x}) + (-a+c)hu^{\prime}(\overline{x}) \\ 
			&+ \frac{1}{2}(a+c)h^{2}u^{\prime\prime}(\overline{x}) + \frac{1}{6}(-a+c)h^{3}u^{\prime\prime\prime}(\overline{x}) + O(h^{4})
		\end{aligned}
	\end{equation}
	求解线性方程组
	\begin{equation}\label{key}
		\left\{
		\begin{aligned}
			&a+b+c = 0\\ 
			&-a+c = 0\\ 
			&a+c = \frac{2}{h^{2}}
		\end{aligned}
		\right.
	\end{equation}
	得到$ a = c = \frac{1}{h^{2}},b = -\frac{2}{h^{2}} $，从而有
	\begin{equation}\label{key}
		D^{2}u(\overline{x}) = \dfrac{u(\overline{x}-h) - 2u(\overline{x}) + u(\overline{x}+h)}{h^{2}}
	\end{equation}
	注意到$ -a + c = 0 $，则3阶项为0，而由4阶展开的 Lagrange 余项
	\begin{equation}\label{key}
		\dfrac{1}{24}(a+c)h^{4}f^{(4)}(\xi) = \dfrac{h^{2}}{12}f^{(4)}(\xi),\ \xi\in[\overline{x}-h,\overline{x}+h]
	\end{equation}
	再由每一项的扰动$ \epsilon\in[-E,E] $
	\begin{equation}\label{key}
		\left|\Delta D^{2}u(\overline{x})\right|\le \dfrac{\left|\Delta u(\overline{x}-h)\right| + 2\left|\Delta u(\overline{x})\right| + \left|\Delta u(\overline{x}+h)\right|}{h^{2}} = \dfrac{4E}{h^{2}}
	\end{equation}
	从而有
	\begin{equation}\label{key}
		\left|u^{\prime\prime}(\overline{x})-D^{2}u(\overline{x})\right|\le\dfrac{h^{2}}{12}\left|f^{(4)}(\xi)\right| + \dfrac{4E}{h^{2}}
	\end{equation}
	由均值不等式有
	\begin{equation}\label{key}
		\dfrac{h^{2}}{12}\left|f^{(4)}(\xi)\right| + \dfrac{4E}{h^{2}}\ge \sqrt{\dfrac{4}{3}E\left|f^{(4)}(\xi)\right|}
	\end{equation}
	其中等式成立当且仅当
	\begin{equation}\label{key}
		h = \sqrt[4]{\dfrac{48E}{\left|f^{(4)}(\xi)\right|}}
	\end{equation}
	此时误差界最小.\\ 
	\indent 设有 FD 公式
	\begin{equation}\label{key}
		D^{4}u(\overline{x}) = au(\overline{x}-2h) + bu(\overline{x}-h) + cu(\overline{x}) + du(\overline{x}+h) + eu(\overline{x}+2h)
	\end{equation}
	在$ x = \overline{x} $处 Taylor 展开得
	\begin{equation}\label{key}
		\begin{aligned}
			D^{4}u(\overline{x}) &= (a+b+c+d+e)u(\overline{x}) + (-2a-b+d+2e)hu^{\prime}(\overline{x})\\ 
			&+ \dfrac{1}{2}(4a+b+d+4e)h^{2}u^{\prime\prime}(\overline{x}) + \dfrac{1}{6}(-8a-b+d+8e)h^{3}u^{\prime\prime\prime}(\overline{x})\\ 
			&+ \dfrac{1}{24}(16a+b+d+16e)h^{4}u^{(4)}(\overline{x}) + O(h^{5})
		\end{aligned}
	\end{equation}
	从而有线性方程组
	\begin{equation}\label{key}
		\left\{
		\begin{aligned}
			&a+b+c+d+e = 0\\ 
			&-2a-b+d+2e = 0\\ 
			&4a+b+d+4e = \frac{2}{h^{2}}\\ 
			&-8a-b+d+8e = 0\\ 
			&16a+b+d+16e = 0
		\end{aligned}
		\right.
	\end{equation}
	解得
	\begin{equation}\label{key}
		a = -\dfrac{1}{12h^{2}},\ b = \dfrac{4}{3h^{2}},\ c = -\dfrac{5}{2h^{2}},\ d = \dfrac{4}{3h^{2}},\ e = -\dfrac{1}{12h^{2}}
	\end{equation}
	从而有
	\begin{equation}\label{key}
		D^{4}u(\overline{x}) = \dfrac{-u(\overline{x}-2h)+16u(\overline{x}-h)-30u(\overline{x})+16u(\overline{x}+h)-u(\overline{x}+2h)}{12h^{2}}
	\end{equation}
	注意到5阶余项为0，而6阶余项为
	\begin{equation}\label{key}
		\dfrac{h^{6}}{6!}(64a+b+d+64e)f^{(6)}(\xi) = -\dfrac{h^{4}}{90}f^{(6)}(\xi),\ \xi\in[\overline{x}-2h,\overline{x}+2h]
	\end{equation}
	因此它是4阶精确的，并且
	\begin{equation}\label{key}
		\left|u^{\prime\prime}(\overline{x})-D^{4}u(\overline{x})\right|\le\dfrac{h^{4}}{90}\left|f^{(6)}(\xi)\right| + \dfrac{16E}{3h^{2}}
	\end{equation}
	由均值不等式有
	\begin{equation}\label{key}
		\dfrac{h^{4}}{90}\left|f^{(6)}(\xi)\right| + \dfrac{8E}{3h^{2}} + \dfrac{8E}{3h^{2}}\ge 4\sqrt[3]{\dfrac{E^{2}\left|f^{(6)}(\xi)\right|}{30}}
	\end{equation}
	其中等式成立当且仅当
	\begin{equation}\label{key}
		h = \sqrt[6]{\dfrac{240E}{\left|f^{(6)}(\xi)\right|}}
	\end{equation}
	此时误差界最小.\\ 
	\indent 比较两种情况，可以发现公式系数关于$ \overline{x} $对称，并且符号逐项交替，系数从两边向中间逐渐增大，$ u(\overline{x}) $的系数总是负的；作差可得
	\begin{equation}\label{key}
		D^{4}u(\overline{x}) - D^{2}u(\overline{x}) = \dfrac{1}{12h^{2}}[-u(\overline{x}-2h)+4u(\overline{x}-h)-6u(\overline{x})+4u(\overline{x}+h)-u(\overline{x}+2h)]
	\end{equation}
	我们发现其中的系数恰为4次多项式的二项式系数，相对应的
	\begin{equation}\label{key}
		D^{2}u(\overline{x}) = \dfrac{1}{h^{2}}[u(\overline{x}-h) - 2u(\overline{x}) + u(\overline{x}+h)]
	\end{equation}
	其中的系数恰为2次多项式的二项式系数.

	\section*{Exercise 6.41}
	构造 FD 公式
	\begin{equation}\label{key}
		D_{+}u(\overline{x}) = au(\overline{x}+h) + bu(\overline{x})
	\end{equation}
	在$ x = \overline{x} $处 Taylor 展开得
	\begin{equation}\label{key}
		D_{+}u(\overline{x}) = (a + b)u(\overline{x}) + ahu^{\prime}(\overline{x}) + \dfrac{1}{2}ah^{2}u^{\prime\prime}(\xi),\ \xi\in[\overline{x}-h,\overline{x}+h]
	\end{equation}
	求解线性方程组
	\begin{equation}\label{key}
		\left\{
		\begin{aligned}
			&a+b = 0\\ 
			&a = \dfrac{1}{h}
		\end{aligned}
		\right.
	\end{equation}
	得到$ a = \frac{1}{h}, b = -\frac{1}{h} $，从而有2阶余项为
	\begin{equation}\label{key}
		\dfrac{1}{2}ah^{2}u^{\prime\prime}(\xi) = \dfrac{1}{2}hu^{\prime\prime}(\xi)
	\end{equation}
	因此$ D_{+}u(\overline{x}) $是1阶精确的.\\ 
	\indent 类似地，
	\begin{equation}\label{key}
		D_{-}u(\overline{x}) = au(\overline{x}) + bu(\overline{x}-h)
	\end{equation}
	其中$ a = \frac{1}{h}, b = -\frac{1}{h} $，有2阶余项为
	\begin{equation}\label{key}
		\dfrac{1}{2}bh^{2}u^{\prime\prime}(\xi) = -\dfrac{1}{2}hu^{\prime\prime}(\zeta),\ \zeta\in[\overline{x}-h,\overline{x}+h]
	\end{equation}
	对于$ D_0u(\overline{x}) $有
	\begin{equation}\label{key}
		D_0u(\overline{x}) = au(\overline{x}+h) + bu(\overline{x}-h)
	\end{equation}
	在$ x = \overline{x} $处 Taylor 展开得
	\begin{equation}\label{key}
		D_{0}u(\overline{x}) = (a + b)u(\overline{x}) + (a-b)hu^{\prime}(\overline{x}) + \dfrac{1}{2}(a+b)h^{2}u^{\prime\prime}(\overline{x}) + O(h^{3})
	\end{equation}
	其中$ a = \frac{1}{2h}, b = -\frac{1}{2h} $，有3阶余项为
	\begin{equation}\label{key}
		\dfrac{1}{6}(a-b)h^{3}u^{\prime\prime\prime}(\xi) = \dfrac{h^{2}}{6}u^{\prime\prime\prime}(\xi),\ \xi\in[\overline{x}-h,\overline{x}+h]
	\end{equation}
	因此$ D_0u(\overline{x}) $是2阶精确的.
	
	\section*{Exercise 6.42}
	计算差分表得
	\begin{center}
		\begin{tabular}{c|ccc}
			$ \overline{x} $ & $ u(\overline{x}) $ & & \\ 
			$ \overline{x}-h $ & $ u(\overline{x}-h) $ & $ \frac{u(\overline{x})-u(\overline{x}-h)}{h} $ & \\ 
			$ \overline{x}-2h $ & $ u(\overline{x}-2h) $ & $ \frac{u(\overline{x}-h)-u(\overline{x}-2h)}{h} $ & $ \frac{u(\overline{x})-2u(\overline{x}-h)+u(\overline{x}-2h)}{2h^{2}} $
		\end{tabular}
	\end{center}
	得到插值多项式
	\begin{equation}\label{key}
		p(x) = u(\overline{x}) + \dfrac{u(\overline{x})-u(\overline{x}-h)}{h}(x-\overline{x}) + \dfrac{u(\overline{x})-2u(\overline{x}-h)+u(\overline{x}-2h)}{2h^{2}}(x-\overline{x})(x-\overline{x}+h)
	\end{equation}
	求导得到
	\begin{equation}\label{key}
		p^{\prime}(x) = \dfrac{u(\overline{x})-u(\overline{x}-h)}{h} + \dfrac{u(\overline{x})-2u(\overline{x}-h)+u(\overline{x}-2h)}{2h^{2}}(2x-2\overline{x}+h)
	\end{equation}
	代入$ x = \overline{x} $有
	\begin{equation}\label{key}
		\begin{aligned}
			p^{\prime}(\overline{x}) &= \dfrac{u(\overline{x})-u(\overline{x}-h)}{h} + \dfrac{u(\overline{x})-2u(\overline{x}-h)+u(\overline{x}-2h)}{2h}\\ 
			&= \dfrac{3u(\overline{x})-4u(\overline{x}-h)+u(\overline{x}-2h)}{2h}
		\end{aligned}
	\end{equation}
	
\end{document}